03_InstSolManual_PDF_Part18

03_InstSolManual_PDF_Part18 - x t 2 v 5" 2 gh sin u...

This preview shows page 1. Sign up to view the full content.

3.60. Set Up: Let be downward. Solve: Use the vertical motion to find the time in the air: with gives The quadratic formula gives The positive root is Then 3.61. Set Up: Example 3.5 derives Solve: (for 3.62. Set Up: Figure 3.43 in the textbook gives the horizontal range to be and the maximum height to be Example 3.5 derives and Solve: (a) and (b) and (c) gives 3.63. Set Up: Use coordinates with the origin at the ground and upward. The shot put has and Solve: gives and gives Use and solve for t . This gives Then Reflect: Since the initial and final heights are not the same, the expression does not apply. 3.64. Set Up: For circular motion the acceleration has magnitude and is directed toward the center of the circle. The period T (time for one revolution) is T 5 2 p R v . a rad 5 v 2 R R 5 v 0 2 sin 1 2 u 0 2 g v 0 5 30.17 m 2.09 s 5 14.4 m / s 5 32.2 mph t 5 2.09 s. v 0 t 5 30.17 m 0 5 2.00 m 1 1 v 0 sin 40.0° 2 t 2 1 4.90 m / s 2 2 t 2 . y 2 y 0 5 v 0 y t 1 1 2 a y t 2 v 0 t 5 30.17 m 23.11 m 5 1 v 0 cos 40.0° 2 t x 2 x 0 5 v 0 x t 1 1 2 a x t 2 1 mph 5 0.4470 m / s a y 52 9.80 m / s 2 . a x 5 0 v 0 y 5 v 0 sin u 0 , v 0 x 5 v 0 cos u 0 , y 0 5 2.00 m, 1 y t 5 x 2 x 0 v 0 x 5 R v 0 cos u 0 5 25.0 m 1 15.9 m / s 2 cos 38.1° 5 2.00 s x 2 x 0 5 v 0 x t 1 1 2 a
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x t 2 v 5 " 2 gh sin u 5 " 2 1 9.80 m / s 2 21 4.90 m 2 sin 38.1° 5 15.9 m / s u 5 38.1° tan u 5 4 h R 5 4 1 4.90 m 2 25.0 m h R 5 1 v 2 sin 2 u 2 g 21 g 2 v 2 sin u cos u 2 5 tan u 4 h 5 v 2 sin 2 u 2 g . R 5 2 v 2 sin u cos u g h 5 4.90 m. R 5 25.0 m v 5 " R max g 5 "1 7.52 m 21 9.80 m / s 2 2 5 8.58 m / s 5 19.2 mph u 5 45°). R max 5 v 2 g 1 mph 5 0.4470 m / s R 5 v 2 sin 1 2 u 2 g . x 2 x 5 v x t 1 1 2 a x t 2 5 1 5.36 m / s 21 1.29 s 2 5 6.91 m. t 5 1.29 s. t 5 1 2 1 4.9 2 1 2 4.50 6 "1 4.50 2 2 2 4 1 4.9 21 2 14.0 2 2 s. 1 4.50 m / s 2 t 1 1 4.9 m / s 2 2 t 2 . 14.0 m 5 y 2 y 5 14.0 m y 2 y 5 v y t 1 1 2 a y t 2 4.50 m / s. v y 5 v sin u 5 v x 5 v cos u 5 5.36 m / s, a y 5 1 9.80 m / s 2 . a x 5 0, 1 y 3-18 Chapter 3...
View Full Document

This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

Ask a homework question - tutors are online